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\title{\heiti\zihao{2} 习题3.1}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{用$\varepsilon-\delta$ 定义证明下列极限:}
\subsection{$\lim _{x \rightarrow 2}\left(x^{2}-6 x+10\right)=2$}
\textbf{证}\quad
$\forall \varepsilon>0$, 要证 $\left|\left(x^{2}-6 x+10\right)-2\right|=|x-2| \cdot|x-4|<\varepsilon$, 当 $x \rightarrow 2$ 时,要对 $|x-4|$ 做
出估计, 所以考虑 $x$ 在 $2$ 的一个小范围内变化.假定 $|x-2|<1$, 则 $1<x<3 \Rightarrow-3<x-4<-1$, 故 $1<|x-4|<3$, 此时
$$
	\left|\left(x^{2}-6 x+10\right)-2\right|=|x-2| \cdot|x-4|<3|x-2|<\varepsilon
$$

所以 $\forall \varepsilon>0$, 取$ \delta=\min \left\{1, \frac{\varepsilon}{3}\right\}$, 则当 $0<|x-2|<\delta$ 时 ,$\left|\left(x^{2}-6 x+10\right)-2\right|<\varepsilon$.故
$$
	\lim _{x \rightarrow 2}\left(x^{2}-6 x+10\right)=2
$$

\subsection{$\lim _{x \rightarrow 1} \frac{x-1}{x^{2}-1}=\frac{1}{2}$}
\textbf{证}\quad
$\forall \varepsilon>0$,要证$ \left|\frac{x-1}{x^{2}-1}-\frac{1}{2}\right|<\varepsilon$, 因为
$$
	\left|\frac{x-1}{x^{2}-1}-\frac{1}{2}\right|=\left|\frac{1}{x+1}-\frac{1}{2}\right|=\left|\frac{1-x}{2(x+1)}\right|=\left|\frac{x-1}{2(x+1)}\right|
$$

当 $x \rightarrow 1$  时,要对 $|x+1|$ 做出估计,所以考虑 $x$ 在 $1$ 的一个小范围内变化,假定 $|x-1|<1$, 则
$0<x<2 \Rightarrow 1<x+1<3$, 故$ 1<|x+1|<3$, 此时
$$
	\left|\frac{x-1}{x^{2}-1}-\frac{1}{2}\right|=\left|\frac{x-1}{2(x+1)}\right|<\frac{|x-1|}{2}<\varepsilon
$$

所以 $\forall \varepsilon>0 取 \delta=\min \{1,2 \varepsilon\}$, 则当 $0<|x-1|<\delta$ 时 ,$\left|\frac{x-1}{x^{2}-1}-\frac{1}{2}\right|<\varepsilon$, 故
$$
	\lim _{x \rightarrow 1} \frac{x-1}{x^{2}-1}=\frac{1}{2}
$$

\subsection{$\lim _{x \rightarrow x_{0}} \cos x=\cos x_{0}$}
\textbf{证}\quad
要证$\left|\cos x-\cos x_{0}\right|<\varepsilon$,  因为$\left|\cos x-\cos x_{0}\right|=2\left|\sin \frac{x+x_{0}}{2} \sin \frac{x-x_{0}}{2}\right| \leqslant 2\left|\sin \frac{x-x_{0}}{2}\right| \leqslant\left|x-x_{0}\right|<\varepsilon$, 所以  $\forall \varepsilon>0$, 取 $\delta=\varepsilon$, 则当 $ 0<\left|x-x_{0}\right|<\delta$时,$\left|\cos x-\cos x_{0}\right|<\varepsilon$, 故$\lim _{x \rightarrow x_{0}} \cos x=\cos x_{0}$

\section{设 $\lim _{x \rightarrow x_{0}^{+}} f(x)=A$, $\lim _{x \rightarrow x_{0}^{+}} g(x)=B$,若 $A>B$,  证明存在$\delta>0$,使得当$0<x-x_{0}<\delta$ 时,成立$f(x)>g(x)$.}
\textbf{证}\quad
任意$0<\varepsilon<A-B$,$\exists \delta_{1}$, 使得当$0<x-x_{0}<\delta_{1}$时,  $| f(x)-f(x_{0}+0)|<\frac{\varepsilon}{2}$.$\exists \delta_{2}$, 使得当$0<x-x_{0}<\delta_{2}$时, $| g(x)-g(x_{0}+0)|<\frac{\varepsilon}{2}$.取$\min \left\{\delta_{1}, \delta_{2}\right\}=\delta$, 当 $0<x-x_{0}<\delta $时,有$f(x_{0}+0)-f(x)+g(x)-g(x_{0}+0)<\varepsilon<A-B$.故此时$f(x)>g(x)$

\section{已知$\lim _{x \rightarrow x_{0}}f(x)=A$,证明:$\lim _{x \rightarrow x_{0}}|f(x)|=|A|$.}
\textbf{证}\quad
$\forall \varepsilon>0, \exists \delta, \mathrm{s.t.}$若$x \in U_{0}\left(x_{0}, \delta\right)$,则$|f(x)-A|<\varepsilon,\therefore|| f(x)|-| A|| \leqslant|f(x)-A|<\varepsilon$.

故$\lim _{x \rightarrow x_{0}}|f(x)|=|A|$.

\section{证明$\lim _{x \rightarrow x_{0}} f(x)=0$ 等价于 $\lim _{x \rightarrow x_{0}}|f(x)|=0$.}
\textbf{证}\quad
利用上题结论即可.

\section{ 证明函数极限的运算性质 : 若$ \lim _{x \rightarrow x_{0}} f(x)=A$, $\lim _{x \rightarrow x_{0}} g(x)=B \neq 0$, 则  $\lim _{x \rightarrow x_{0}} \frac{f(x)}{g(x)}=\frac{A}{B}$ .}
\textbf{证}\quad
由海涅定理,对于$\lim _{x \rightarrow x_{0}} f(x)=A$, $\lim _{x \rightarrow x_{0}} g(x)=B \neq 0$,有任意收敛于$x_{0}$的数列$x_{n}$都满足
$$
	\lim _{n \rightarrow\infty } f(x_{n})=A,\lim _{n \rightarrow\infty } g(x_{n})=B
$$

由数列极限的四则运算知$\lim _{n \rightarrow\infty }\frac {f(x_n)}{g(x_{n})}= \frac{A}{B}$,所以由海涅定理,
$$
	\lim _{x \rightarrow x_{0}} \frac{f(x)}{g(x)}=\frac{A}{B}
$$.

\section{使用夹逼定理求下列极限 :}
\subsection{$\lim _{x \rightarrow 0^{+}} x\left[\frac{1}{x}\right]$}
\textbf{解}\quad


\subsection{$\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}\right)^{x}$}
\textbf{解}\quad
$\forall x \rightarrow 0^{+}$, $\exists n \in \mathbb{Z}^{+}\mathrm{s.t.}$
$$
	\frac{1}{1+n}<x \leqslant \frac{1}{n} \Leftrightarrow n \leqslant\left[\frac{1}{x}\right]<1+n
$$

即
$$
	\left[\frac{1}{x}\right]=n.\frac{1}{1+n} \cdot n=\frac{1}{1+n}\cdot\left[\frac{1}{x}\right]<x\left[\frac{1}{x}\right] \leqslant 1
$$

由 $\lim _{n \rightarrow \infty} \frac{n}{n+1}=1$知
$$
	\lim _{x \rightarrow 0^{+}} x\left[\frac{1}{x}\right]=1
$$

$$
	\forall x \rightarrow 0^{-}, \exists n \in \mathbb{Z}^{+}, \mathrm{s.t.}-\frac{1}{n}<x \leqslant-\frac{1}{1+n}
$$

即
$$
	\left[\frac{1}{x}\right]=-(n+1), \quad-\frac{1}{n}[-(n+1)]=-\frac{1}{n}\left[\frac{1}{x}\right]>x\left[\frac{1}{x}\right] \geqslant1
$$

又由
$$
	\lim _{n \rightarrow \infty} \frac{n+1}{n}=1
$$

知
$$
	\lim _{x \rightarrow 0^{-}} x\left[\frac{1}{x}\right]=1\quad\therefore \lim _{x \rightarrow 0} x\left[\frac{1}{x}\right]=1
$$

\subsection{$\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}\right)^{x}$}
\textbf{解}\quad
记$t=\frac{1}{x}$,$\therefore$结论等价于求$\lim _{t \rightarrow \infty}t^{\frac{1}{t}}$
$\because [t]^{\frac{1}{[t]+1}}<t^{\frac{1}{t}}<(2[t])^{\frac{1}{[t]}}$,$\therefore$ 由夹逼定理和已知极限,$\lim _{t \rightarrow \infty}[t]^{\frac{1}{[t]+1}} =\lim _{t \rightarrow \infty}t^{\frac{1}{t}}=\lim _{t \rightarrow \infty}(2[t])^{\frac{1}{[t]}}=1$
$$
	\therefore\lim _{x \rightarrow 0^{+}} (\frac{1}{x})^{x}=1
$$

\section{计算下列极限:}
\subsection{$\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{2}+x-2}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{2}+x-2}=\lim _{x \rightarrow 1}\frac{(x+1)(x-1)}{(x-1)(x+2)}=\lim _{x \rightarrow 1} \frac{x+1}{x+2}=\frac{2}{3}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{(x+1)^{10}-1}{x}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{(x+1)^{10}-1}{x}=\frac{x^{10}+c_{10}^{1} x^{9}+c_{10}^{2} x^{8}+\cdots+c_{10}^{8} x}{x}=\frac{x^{9}+c_{10}^{1} x^{8}+-+c_{10}^{9}}{1}=10
$$

\subsection{$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\sqrt[3]{x}-1}$}
\textbf{解}\quad
记$\sqrt[6]{x}=t$,则
$$
	\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{ \sqrt[3]{x}-1}=\lim _{t \rightarrow 1}\frac{t^{3}-1}{t^{2}-1}=\lim _{t \rightarrow 1}\frac{(t^{2}+t+1)(t-1)}{(t+1)(t-1)}=\frac{3}{2}
$$

\subsection{$\lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7}=\lim _{x \rightarrow 7}\frac{x-7}{(x-7)(\sqrt{x+2}+3)}=\frac{1}{\sqrt{7+2}+3}=\frac{1}{6}
$$

\subsection{$\lim _{x \rightarrow 1} \frac{x+x^{2}+\cdots+x^{k}-k}{x-1}(k \in \mathbb{N}^{*})$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 1} \frac{x+x^{2}+\cdots+x^{k}-k}{x-1}=\lim _{x \rightarrow 1}\frac{(x-1)+(x+1)(x-1)+\cdots+(x^{k-1}+\cdots+1)(x-1)}{x-1}=1+2+\cdots +k=\frac{k(k+1)}{2}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\sqrt{x+1}-\sqrt[3]{x+1}}{x}$}
\textbf{解}\quad
记$\sqrt[6]{x+1}=t$,则
$$
	\lim _{x \rightarrow 0} \frac{\sqrt{x+1}-\sqrt[3]{x+1}}{x}=\lim _{t \rightarrow 1}\frac{t^{3}-t^{2}}{t^{6}-1}=\frac{t^{2}(t-1)}{(t^{5}+t^{4}+t^{3}+t^{2}+t+1)(t-1)}=\frac{1}{6}
$$

\subsection{$\lim _{x \rightarrow 1} (\frac{k}{x^{k}-1}-\frac{l}{x^{l}-1}), k, l \in \mathbb{N}^{*}$}
\textbf{解}\quad
记$t+1=x,g(x)=(x^{k-1}+-+1)(x^{l-1}+\cdots+1)$,则
$$
	\begin{aligned}
		\lim _{x \rightarrow 1} (\frac{k}{x^{k}-1}-\frac{l}{x^{l}-1}) & =\lim _{x \rightarrow 1}\frac{k(x^{l}-1)-l( x^{k}-1)}{(x^{k}-1)(x^{l}-1)} \\&=\lim _{t \rightarrow 0} \frac{k[(t+1)^{l}-1]-l[(t+1)^{k}-1]}{g(t+1) t^{2}}
		\\&=\lim _{t \rightarrow 0} \frac{k[1+l t+\frac{l(l+1)}{2} t^{2}+\cdots-1]-l[1+k t+\frac{k(k+1)}{2} t^{2}+\cdots-1]}{g(t+1) t^{2}}
		\\&=\lim _{t \rightarrow 0} \frac{(k l \cdot \frac{l+1}{2}-k l \frac{k+1}{2}) t^{2}+o(t^{2})}{g(t+1) t^{2}} \\&= \frac{(k l \cdot \frac{l+1}{2}-k l \frac{k+1}{2})}{g(1)}\\&=\frac{l+1}{2}-\frac{k+1}{2}\\&=\frac{l-k}{2}
	\end{aligned}
$$

\section{设 $\lim _{x \rightarrow x_{0}} f(x)=A$, $f(x)>0$, 证明 $\lim _{x \rightarrow x_{0}} \sqrt[n]{f(x)}=\sqrt[n]{A}$,其中$n \geqslant 2$为正整数.}
\textbf{证}\quad
$f(x)=\sqrt[n]{x}$在定义域内任意一点连续(在$x=0$时右连续),所以由复合函数的极限定理显然.

\section{已知 $f(x)=\left\{\begin{array}{ll}\cos x, & x>0 \\ 1 & x=0, \text { 求 } f(x) \text { 在 } x=0 \text { 的左右极限. } \\ x^{2}+2, & x<0\end{array}\right.$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & f(x+0)=\lim _{x \rightarrow 0+0}cos(x)=1
		\\&f(x-0)=\lim _{x \rightarrow 0-0}x^2+2=2
	\end{aligned}
$$
所以左极限为$2$,右极限为$1$.

\section{设函数$f(u)=\{\begin{array}{ll}
		  1, & u=0,      \\
		  0, & u \neq 0,
	  \end{array} u=g(x)=x \sin \frac{1}{x}$,证明 : $\lim _{x \rightarrow 0} f(g(x))$ 不存在.}
\textbf{证}\quad
$x_{n}=\frac{1}{2n\pi},f(g(x_{n}))$当$n\rightarrow\infty$时，由于$g(x)$为$0$显然其极限为$1$.

而当$x$为$\frac{1}{(2n+1)\pi}$时,$|x \sin \frac{1}{x}|=|x|$当$x\rightarrow 0$时,$g(x)$为$1$,$f(g(x))$极限为$0$

两个收敛到$0$的子列$x_{n_{1}}$和$x_{n_{2}}$的函数值的数列(原式的子列)极限不相等,所以极限不存在.

\section{计算下列极限:}
\subsection{$\lim _{x \rightarrow 0} \frac{\sin 3 x-\sin x}{\sin 2 x}$}
\textbf{解}\quad
由函数极限的四则运算
$$
	\lim _{x \rightarrow 0} \frac{\sin 3 x-\sin x}{ \sin 2 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{x}-\frac{\sin x}{x}}{\frac{\sin 2 x}{x}}=\lim _{x \rightarrow 0} \frac{3 \cdot \frac{\sin 3 x}{3 x}-\frac{\sin x}{x}}{2-\frac{\sin  2 x}{2 x}}=1
$$

\subsection{$\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos x}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos x}=\frac{x^{2}}{2 \sin ^{2} \frac{x}{2}}=2
$$

\subsection{$\lim _{x \rightarrow 0} \frac{1-\cos x \cdot \cos 3 x}{x^{2}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} \frac{1-\cos x \cos 3 x}{x^{2}} & =\lim _{x \rightarrow 0} \frac{1-(1-\sin ^{2} \frac{x}{2})(1-2 \sin ^{2} \frac{3 x}{2})}{x^{2}} \\&=\lim _{x \rightarrow 0} \frac{1-1+2 \sin ^{2} \frac{x}{2}+2 \sin ^{2} \frac{3 x}{2}-4 \sin ^{2} \frac{x}{2} \cdot \sin ^{2} \frac{3 x}{2}}{x^{2}}
		\\&=\frac{1}{2}+\frac{9}{2}
		\\&=5
	\end{aligned}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\cos x-\cos 3 x}{2 x^{2}}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{\cos x-\cos 3 x}{2 x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin x \sin 2 x}{2x^{2}}= \lim _{x \rightarrow 0} \frac{\sin x}{x} \frac{\sin 2 x}{2 x} \cdot2=2
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\sin (\sin x)}{x}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{\sin (\sin x)}{x}=\lim _{x \rightarrow 0}\frac{\sin (\sin x)}{\sin x}=\lim _{x \rightarrow 0}\frac{\sin x}{x}=1
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\tan (\tan x)}{x}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} \frac{\tan (\tan x)}{x} & =\lim _{x \rightarrow 0}\frac{\tan (\tan x)}{\tan x} \cdot \frac{\tan x}{x} \\&=\lim _{x \rightarrow 0}\frac{\tan x}{x} \cdot \frac{\tan x}{x}\\&=\lim _{x \rightarrow 0}\frac{\sin ^{2} x}{x^{2}} \cdot \frac{1}{\cos ^{2} x}\\&=1
	\end{aligned}
$$

\subsection{$\lim _{n \rightarrow \infty} \frac{2^{n}}{x} \sin \frac{x}{2^{n}} \quad(x \neq 0)$}
\textbf{解}\quad
记$\frac{x}{2^{n}}=t$,则原式:
$$
	\lim _{t \rightarrow 0} \frac{1}{t} \sin t =1
$$

\subsection{$\lim _{x \rightarrow 1}(x-1) \tan \frac{\pi x}{2}$}
\textbf{解}\quad
记$t=x-1$,则原式:
$$
	\begin{aligned}
		\lim _{x \rightarrow 1}(x-1) \tan \frac{\pi x}{2} & =\lim _{x \rightarrow 1}(x-1) \cdot \frac{\sin \frac{\pi x}{2}}{\cos \frac{\pi x}{2}} \\&=\lim _{x \rightarrow 1}\sin \frac{\pi x}{2} \cdot \frac{x-1}{\cos \frac{\pi x}{2}}
		\\&=1\cdot \frac{x-1}{\sin (\frac{\pi}{2}-\frac{\pi}{2}x)}\\&=\lim _{t \rightarrow 0} \frac{t}{\sin \frac{\pi}{2} (-t)}\\&=-\frac{2}{\pi}
	\end{aligned}
$$

\section{求极限$\lim _{n \rightarrow \infty} \sin (\pi \sqrt{n^{2}+\sqrt{n}})$ 和$ \lim _{n \rightarrow \infty} 2 n \sin (\pi \sqrt{4 n^{2}+1})$ , $n$ 为正整数.}
\subsection{}
\textbf{解}\quad
先证$\lim _{n \rightarrow \infty} \sqrt{n^{2}+\sqrt{n}}=n$

$\lim _{n \rightarrow \infty} \sqrt{n^{2}+\sqrt{n}}-n=\lim _{n \rightarrow \infty}\frac{\sqrt{n}}{\sqrt{n^{2}+\sqrt{n}}+n}=0$,所以得证.
$$
	\begin{aligned}
		\lim _{n \rightarrow \infty} \sin (\pi \cdot \sqrt{n^{2}+\sqrt{n}}) \\&=\lim _{n \rightarrow \infty} \sin [\pi \cdot(\sqrt{n^{2}+n}-n) +n \pi]
		\\&=\lim _{n \rightarrow \infty} \sin [\pi(\sqrt{n^{2}+n}-n)] \cos n \pi+\cos (\pi (\sqrt{n^{2}+n}-n)) \sin n \pi\\&=0+0=0
	\end{aligned}
$$

($\sin x $在$x\rightarrow 0$时趋近于$0$)

\subsection{}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{n \rightarrow \infty} 2 n \sin (\pi \cdot \sqrt{4 n^{2}+1}) & =\lim _{n \rightarrow \infty} 2 n \sin [\pi(\sqrt{4 n^{2}+1}-2 n)]                                                                                         \\
		                                                                   & =\lim _{n \rightarrow \infty} 2 n \sin [\pi \cdot \frac{1}{\sqrt{4 n^{2}+1}+2 n}]                                                                          \\
		                                                                   & =\lim _{n \rightarrow \infty} \frac{\sin (\frac{\pi}{\sqrt{4 n^{2}+1}+2 n})}{\frac{\pi}{\sqrt{4 n^{2}+1}+2 n}} \cdot \frac{2 n \pi}{\sqrt{4 n^{2} 41}+2 n} \\
		                                                                   & =\frac{\pi}{2}
	\end{aligned}
$$

\section{证明 $\lim _{x \rightarrow 0}\left[\lim _{n \rightarrow \infty}\left(\cos x \cos \frac{x}{2} \cdot \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n}}\right)\right]=1, n$ 为正整数.}
\textbf{证}\quad
$$
	\begin{aligned}
		\cos x \cos \frac{x}{2} \cdots \cos \frac{x}{2^{n}}
		 & =\frac{\sin \frac{x}{2^{n}} \cdot \cos \frac{x}{2^{n}} \cos \frac{x}{2^{n-1}} \cdot \cos \frac{x}{2^{n-2}} \cdot \cdots \cdot \cos x}{\sin \frac{x}{2^{n}}} \\
		 & =\frac{\sin \frac{x}{2^{n-1}} \cos \frac{x}{2^{n-1}} \cdot \cos \frac{x}{2^{n-2}} \cdots \cos x}{2 \sin \frac{x}{2^{n}}}                                    \\
		 & =\frac{\sin 2 x}{2^{n+1} \sin \frac{x}{2^{n}}}
	\end{aligned}
$$

故
$$
	\begin{aligned}
		\lim _{x \rightarrow 0}[\lim _{n \rightarrow \infty}(\cos x \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n}})] & =\lim _{x \rightarrow 0}(\lim _{n \rightarrow \infty} \frac{\sin 2 x}{2^{n+1} \sin \frac{x}{2^{n}}})                         \\
		                                                                                                                            & =\lim _{x \rightarrow 0} \lim _{n \rightarrow \infty} \frac{\sin 2 x}{2 x} \cdot\frac{\frac{x}{2^{n}}}{\sin \frac{x}{2^{n}}} \\&=1
	\end{aligned}
$$

\section{常用代数变形}
$$
	\begin{aligned}
		 & \sin \frac{(2 n+1)}{2} x=2 \sin \frac{x}{2}\left(\frac{1}{2}+\sum_{i=1}^{n} \operatorname{cosin} x\right) \\
		 & \cos \frac{(2 n+1)}{2} x=-2 \sin\frac{x}{2} \left(\sum_{i=1}^{n} \cos i x\right)                          \\
		 & \cos x \cos \frac{x}{2} \cdots \cos \frac{x}{2^{n}}
		=\frac{\sin \frac{x}{2^{n}} \cdot \cos \frac{x}{2^{n}} \cos \frac{x}{2^{n-1}} \cdot \cos \frac{x}{2^{n-2}} \cdot \cdots \cdot \cos x}{\sin \frac{x}{2^{n}}}
	\end{aligned}
$$
\end{document}
